The friction at sleeve A can provide a maximum resisting moment of 125 N•m about the x axis. Determine the largest magnitude of force F that can be applied to the bracket so that the bracket will not turn.

#### Solution:

Show me the final answer↓

Let us first express force F in Cartesian form. (Don’t remember?)

F=\left\{-F\cos60^0i+F\cos60^0j+F\cos45^0k\right\}

(simplify)

F=\left\{-0.5F\,i+0.5F\,j+0.707F\,k\right\}

We will now draw a position vector from A to B as follows:

Let us now calculate r_{AB}:

r_{AB}=\left\{(-0.15-0)i+(0.3-0)j+(0.1-0)k\right\}

r_{AB}=\left\{-0.15i+0.3j+0.1k\right\}

r_{AB}=\left\{-0.15i+0.3j+0.1k\right\}

We can now calculate the moment along the x-axis. Remember that the unit vector for the x-axis is **i**.

M_x=i\cdot r_{AB}\times F

M_x=\begin{bmatrix}1&0&0\\-0.15&0.3&0.1\\-0.5F&0.5F&0.707F\end{bmatrix}

M_x=\begin{bmatrix}1&0&0\\-0.15&0.3&0.1\\-0.5F&0.5F&0.707F\end{bmatrix}

Taking the cross product gives us:

M_x=0.1621F

Substitute the maximum resisting moment (given to us in the question):

125=0.1621F

F=771 N

F=771 N

#### Final Answer:

F=771 N